Core course · Circuit & Network Theory

Circuit & Network Theory

Learn to analyse any electrical circuit — from Kirchhoff's laws to Thevenin's theorem and AC impedance. Clear notes, worked examples, two calculators and a quiz. Useful for diploma, B.Tech and GATE preparation.

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Lesson 1

Circuit elements and sources

Every network is built from a few basic elements. Knowing how each one behaves is the first step in any analysis.

ElementRelationBehaviour
Resistor (R)v = R·iDissipates energy as heat. Same behaviour for DC and AC.
Inductor (L)v = L·di/dtStores energy in a magnetic field and opposes changes in current. Acts as a short circuit at steady DC.
Capacitor (C)i = C·dv/dtStores energy in an electric field and opposes changes in voltage. Acts as an open circuit at steady DC.
Ideal voltage sourcefixed VSame voltage at any current; zero internal resistance. Replace it with a short circuit when switching it off in a theorem.
Ideal current sourcefixed ISame current at any voltage; infinite internal resistance. Replace it with an open circuit when switching it off.

Active and passive

Active elements supply energy (sources, batteries, solar cells). Passive elements only absorb or store it (R, L, C).

Linear and bilateral

A linear element obeys proportionality (R, L, C). A bilateral element behaves the same in both directions (a resistor does; a diode does not). Most network theorems need linear, bilateral circuits.

Lesson 2

Kirchhoff's laws

KCL — current law

The algebraic sum of currents at any node is zero: current in = current out. It follows from conservation of charge.

Σ Iin = Σ Iout

Example: 5 A and 3 A enter a node and 2 A leaves through one branch, so the other branch carries 6 A out.

KVL — voltage law

The algebraic sum of voltages around any closed loop is zero: rises = drops. It follows from conservation of energy.

Σ Vrise = Σ Vdrop

Example: a 12 V source with drops of 5 V and 4 V, so the third element drops 3 V.

Terms you will meet

Node: a point where two or more elements join. Branch: one element or path between two nodes. Loop: any closed path. Mesh: a loop with no other loop inside it.

Lesson 3

Series, parallel and divider rules

  1. Series and parallel resistance

    Series: R = R₁ + R₂ + … Parallel: 1/R = 1/R₁ + 1/R₂ + …
    Two in parallel: R = R₁·R₂ ÷ (R₁ + R₂)

  2. Voltage divider (series resistors)

    V₂ = V × R₂ ÷ (R₁ + R₂)

    Example: 12 V across 4 Ω and 6 Ω in series. Voltage across the 6 Ω = 12 × 6 ÷ 10 = 7.2 V.

  3. Current divider (parallel resistors)

    I₁ = I × R₂ ÷ (R₁ + R₂)

    The larger current flows through the smaller resistance. Example: 10 A into 2 Ω ∥ 3 Ω gives 6 A through the 2 Ω and 4 A through the 3 Ω.

  4. Source transformation

    Voltage source V with series R ⇄ current source I = V ÷ R with parallel R

    Use it to simplify a network step by step before applying other methods.

Lesson 4

Nodal and mesh analysis

Two systematic methods that work for any linear circuit. Both worked examples below use this circuit, and both give the same answer.

10 V+ at top 5 V+ at top 2 Ω 4 Ω 4 Ω Node V Mesh 1 (I₁ ↻)Mesh 2 (I₂ ↻) Reference node (0 V)
Find the node voltage V and every branch current.

  1. Nodal analysis (based on KCL)

    Pick a reference node, write KCL at every other node in terms of node voltages, and solve. The sum of currents leaving node V is zero:

    (V − 10)/2 + V/4 + (V − 5)/4 = 0 → 2V − 20 + V + V − 5 = 0 → V = 6.25 V

    Currents: from the 10 V source (10 − 6.25) ÷ 2 = 1.875 A; down through the middle 4 Ω, 6.25 ÷ 4 = 1.5625 A; into the 5 V source, (6.25 − 5) ÷ 4 = 0.3125 A. Check with KCL: 1.5625 + 0.3125 = 1.875 A ✔

  2. Mesh analysis (based on KVL)

    Give each mesh a clockwise current and write KVL around it. A shared element carries the difference of the two mesh currents.

    Mesh 1: 10 = 2I₁ + 4(I₁ − I₂) → 6I₁ − 4I₂ = 10
    Mesh 2: 4I₂ + 5 + 4(I₂ − I₁) = 0 → −4I₁ + 8I₂ = −5

    Solving: I₁ = 1.875 A, I₂ = 0.3125 A. The middle 4 Ω carries I₁ − I₂ = 1.5625 A downward, so it drops 6.25 V, the same as the nodal answer ✔

  3. Which method to choose?

    Use nodal analysis when there are fewer nodes than meshes, or when there are current sources. Use mesh analysis when there are fewer meshes, or when there are voltage sources. Both work for any planar circuit.

Lesson 5

Network theorems

Shortcuts that turn a complicated network into a simple one, especially when you only care about one load.

  1. Superposition theorem

    In a linear circuit with several sources, any response equals the sum of the responses caused by each source acting alone, with the others switched off (voltage sources shorted, current sources opened).

    Same circuit as above: 10 V alone (5 V shorted) gives V = 5 V; 5 V alone (10 V shorted) gives V = 1.25 V; together V = 5 + 1.25 = 6.25 V ✔. Superposition works for voltage and current, but not for power.

  2. Thevenin's theorem

    Any linear two-terminal network can be replaced by one voltage source Vth in series with one resistance Rth.

    Vth = open-circuit voltage at the terminals Rth = resistance seen with all sources switched off

    Example: 12 V source, 4 Ω in series, 6 Ω across the load terminals. Vth = 12 × 6 ÷ 10 = 7.2 V; Rth = 4 ∥ 6 = 2.4 Ω. For a 6 Ω load: I = 7.2 ÷ (2.4 + 6) = 0.857 A.

  3. Norton's theorem

    The same network can instead be replaced by a current source IN in parallel with RN.

    IN = short-circuit current = Vth ÷ Rth RN = Rth

    For the example above: IN = 7.2 ÷ 2.4 = 3 A, in parallel with 2.4 Ω.

  4. Maximum power transfer theorem

    A load receives the most power when its resistance equals the Thevenin resistance of the source network.

    RL = Rth Pmax = Vth² ÷ (4 Rth)

    For the example: RL = 2.4 Ω gives Pmax = 7.2² ÷ 9.6 = 5.4 W, at only 50% efficiency. Solar inverters use the same idea: MPPT keeps adjusting the operating point to draw maximum power from the panels.

  5. Reciprocity and Millman's theorems

    Reciprocity: in a linear, bilateral network with one source, swapping the source and an ammeter does not change the reading. Millman's: parallel voltage sources with series resistances combine into one: V = Σ(Vk/Rk) ÷ Σ(1/Rk).

Lesson 6

Star–delta (Y–Δ) transformation

Some networks, such as bridges, are neither series nor parallel. Converting three resistors between star and delta form unlocks them.

Delta → Star

RA = RAB·RCA ÷ (RAB + RBC + RCA)

Each star arm = product of the two adjacent delta sides ÷ sum of all three. RB and RC follow the same pattern.

Star → Delta

RAB = (RARB + RBRC + RCRA) ÷ RC

Each delta side = sum of the pairwise products ÷ the opposite star arm.

Balanced case: Rstar = Rdelta ÷ 3 (e.g. 30 Ω delta = 10 Ω star)
Lesson 7

Transients in RC and RL circuits

When a switch operates, voltages and currents change exponentially, not instantly. The time constant τ sets the speed.

RC circuit

τ = R·C vC(t) = V(1 − e−t/τ) (charging)

A capacitor's voltage cannot change instantly. Example: 10 kΩ and 100 µF give τ = 1 s.

RL circuit

τ = L ÷ R iL(t) = (V/R)(1 − e−t/τ)

An inductor's current cannot change instantly. Example: 0.5 H and 10 Ω give τ = 0.05 s.

Time1τ2τ3τ4τ5τ
% of final value reached63.2%86.5%95.0%98.2%99.3%

Rule of thumb: the circuit reaches steady state after about 5 time constants.

Lesson 8

AC circuits: reactance, impedance and resonance

  1. Reactance

    XL = 2πfL XC = 1 ÷ (2πfC)

    Inductive reactance rises with frequency; capacitive reactance falls. At 50 Hz, 0.1 H gives XL = 31.4 Ω and 100 µF gives XC = 31.8 Ω.

  2. Impedance of a series RLC circuit

    Z = √(R² + (XL − XC)²) φ = tan⁻¹((XL − XC) ÷ R) I = V ÷ Z

    If XL > XC the circuit is inductive and current lags voltage; if XC > XL it is capacitive and current leads.

  3. AC power

    P = VI cos φ (W) Q = VI sin φ (VAR) S = VI (VA) S² = P² + Q²

    Power factor = cos φ = R ÷ Z. Capacitor banks improve the PF of inductive loads such as motors.

  4. Series resonance

    f0 = 1 ÷ (2π√(LC)) at f0: XL = XC, Z = R (minimum), current is maximum

    Quality factor Q = (1/R)·√(L/C); bandwidth = f0 ÷ Q. With 0.1 H and 100 µF, f0 ≈ 50.3 Hz, almost exactly mains frequency.

Practice

Series RLC calculator

Change any value — reactance, impedance, current, power and resonance update instantly.

Circuit

Ω
mH
µF
Hz

Supply

V

Results

XL
XC
Impedance Z
Current I
Phase angle
Power factor
Real power P
Resonant f0

Star ↔ delta converter

Enter three resistances and choose the direction.

Convert

Ω
Ω
Ω

Results

Study material

Quick formula sheet

Revise before your exam. Use "Print / save notes as PDF" at the top to keep a copy.

TopicFormula / rule
KCL / KVLΣ I at a node = 0 Σ V around a loop = 0
Voltage dividerV₂ = V·R₂ ÷ (R₁ + R₂)
Current dividerI₁ = I·R₂ ÷ (R₁ + R₂)
Source transformationV, series R ⇄ I = V/R, parallel R
Thevenin / NortonVth = Voc, IN = Isc, Rth = Voc ÷ Isc
Maximum power transferRL = Rth, Pmax = Vth² ÷ 4Rth
Delta → starRA = RABRCA ÷ ΣRΔ
Star → deltaRAB = (RARB + RBRC + RCRA) ÷ RC
Time constantsτRC = RC, τRL = L/R; steady state after ≈ 5τ
ReactanceXL = 2πfL, XC = 1/(2πfC)
Series RLC impedanceZ = √(R² + (XL − XC)²)
Resonancef0 = 1 ÷ 2π√(LC), Q = (1/R)√(L/C)
AC powerP = VI cos φ, Q = VI sin φ, S = VI
Track your progress

Network theory topics checklist

Test yourself

Network theory quiz

1. Currents of 4 A and 6 A enter a node. One branch carries 3 A out. What does the other branch carry out?

KCL: 4 + 6 = 3 + I, so I = 7 A.

2. When finding Rth, an independent voltage source is replaced by…

An ideal voltage source has zero internal resistance, so it becomes a short. Current sources become open circuits.

3. A network has Vth = 20 V and Rth = 5 Ω. What is the maximum power it can deliver to a load?

Pmax = Vth² ÷ 4Rth = 400 ÷ 20 = 20 W, with RL = 5 Ω.

4. A balanced delta has 45 Ω in each side. The equivalent star resistance per arm is…

Balanced case: Rstar = Rdelta ÷ 3 = 15 Ω.

5. An RC circuit has R = 2 kΩ and C = 50 µF. What is its time constant?

τ = RC = 2000 × 50 × 10⁻⁶ = 0.1 s.

6. At series resonance, the impedance of an RLC circuit is…

XL and XC cancel, leaving Z = R, so the current is at its maximum.

7. Superposition can be used directly to find…

Power depends on I², which is not linear, so add the currents first and then calculate the power.

8. In a series RL circuit with R = 3 Ω and XL = 4 Ω, the power factor is…

Z = √(3² + 4²) = 5 Ω; PF = R ÷ Z = 0.6, lagging because the circuit is inductive.

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